Roots are nice, but we prefer dealing with regular numbers as much as possible. So, for example, instead of √4 we prefer dealing with 2. What about roots that aren't equal to an integer, like √20? Still, we can write 20 as 4⋅5 and then use known properties to write √(4⋅5) as √4⋅√5, which is 2√5. We *simplified* √20. Created by Sal Khan.
Want to join the conversation?
- At0:28, you added all the values and observed that if the sum was divisible by 3, so was the value. What video can I find this principle?(180 votes)
- The link is broken, here is the new one:
- At0:09, Sal said that 117 is not a perfect square. What does that mean?(38 votes)
- A perfect square is a square root is not a decimal. You can not take the square root of 117 and have it not be a decimal. But if you were to take the square root of 9, it would be 3 because 3x3=9. Hope this helped!(53 votes)
- I'm having a LOT of trouble siplifying square roots and I can't understand why it's not making any sense to me...
The Square Roots Practice I can finish in about 10 seconds but I'm really hitting a wall with the Simplification side of Square Roots. Please help me!(51 votes)
- Wouldn't the answer to a square root really be positive and negative? For instance, if we wanted the square root of 9, it would be 3 and -3? because 3x3=9 and -3x-3=9?(34 votes)
- Yes, whenever you take square roots, you get two values (one positive and the other negative). But when you take the "principle" square root , you take only the positive value.(43 votes)
- Around2:24, Sal explains that 5*3 and the square root of thirteen is 15 times the square root of thirteen. Why would you multiply the numbers 5 and 3?(28 votes)
- He is trying to simplify it. 5•3•√13 is more complex than 15•√13. The former has 3 steps involved (multiply 5 and 3, find square root of 13, multiply 15 by square root of 13), while the latter only has 2 steps involved (find square root of 13 and multiply by 15).(33 votes)
- Which video (and where) explains why you can add up the digits of a number to see if it's divisible by 3 like at0:25-0:36?(20 votes)
- go to pre- algabra and in the factors and multiples section you will find divisablity tests at the top of the list and it explains the rule for 3 in the first video(26 votes)
- what is the concept of simplifying square roots? I don't understand square roots(17 votes)
- Roots are the inverse operation to powers. So if you take the square root of 6 and then you square it, then you would be left with 6 because the square and the square root cancel out.
Now if you have the square root of 2 plus the square root of 2, you would have 2√2. Notice that it isn't √4. It is actually 2√2 (which is the same as √8).
So the concept of simplifying square roots is like the concept of simplifying other things like exponents, parentheses, etc.(4 votes)
- Okay so how would you do fractions? I'm very confused and my math teacher sped through it so I didn't understand. How would you simplify the sqare root of 35 over 9 (just and example)?(12 votes)
- The thing about a square root of a fraction is that:
sqrt(35/9) = sqrt(35)/sqrt(9)
in other words, the square root of the entire fraction is the same as the square root of the numerator divided by the square root of the denominator. With that in mind, we can simplify the fraction:
As you can see, I left the numerator under the square root, because I can't simplify it, but the square root of 9 is three so I could replace the sqrt(9) in the denominator by 3.
The same rule applies to exponents: e.g. (2/3)^2=(2^2)/(3^2)(11 votes)
- i still don't understand the concept(11 votes)
- I found a website that breaks this concept down as if they were teaching it to kindergarten students XD
Helped me finally understand this!(14 votes)
- I don't understand.(9 votes)
Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. So we can factor 117 as 3 times 39. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can. We would just leave this as 3 times the square root of 26.